Question

5secx + 5tanx=5

Answer 1

5(secx+secx^2 -1)=0?

Answer 2

5(secx+secx^2-1 -1)=0

Answer 3

5secx + 5tanx=5 I mean divide everything by 5, and you get

secx + tanx=1 THEN...


\[\frac{1}{\cos~x}+\frac{\sin~x}{\cos~x}=1\]

\[\frac{1+\sin~x}{\cos~x}=1\]

\[\frac{1+\sin~x}{\cos~x}=\frac{\cos~x}{\cos~x}\]

\[1+\sin~x=\cos~x\]

\[1=\cos(x)-\sin(x)\] true or false ?

Wait, maybe you need to solve it ?

Answer 4

I need to find all the solutions of the equation in the interval [pi, 2pi) algebraically

Answer 5

algebraically ?

Answer 6

idk, all I can suggest is

\[1=\cos(x)-\sin(x)\]
\[1=\sqrt{1-\sin^2(x) }-\sin(x)\]
\[1+\sin~x=\sqrt{1-\sin^2(x) }\]
\[\sin^2x+2\sin~x+1=1-\sin^2(x) \]
\[2\sin^2x+2\sin~x=0\]
\[2(\sin^2x+\sin~x)=0\]
\[\sin^2x+\sin~x=0\]
can you do it from here ?

Answer 7

yes thank you.

Answer 8

\[l et~~~a=\sin~(x)\]\[a^2+a=0\]\[a(a+1)=0\]

\[\sin(x)=0~~~~or~~~~\sin(x)=1\]
for each one (for =0 and for =1) can you give me al the values of x that would work and that would be
\[180 ≤ x <360\]

Answer 9

180,